Answer:
-60q+60r/4
Step-by-step explanation:
Hope this helps
Answer:
Probability that this whole shipment will be accepted is 0.7324.
Step-by-step explanation:
We are given that a company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives.
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 30 components
r = number of success = fewer than 3
p = probability of success which in our question is % rate
of defects, i.e; 6%
<em>LET X = Number of defective components</em>
So, it means X ~ 
Now, Probability that this whole shipment will be accepted only when there are fewer than 3 defectives = P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
=
=
= 0.7324
Therefore, probability that this whole shipment will be accepted is 0.7324.
Answer:
CDE=50 degrees ACD=100,BAC=30 degrees.
you mean : 5/8x + 1/16x = 5/16 +x
<=> 10/16x +1/16x = (5 +16x)/16
(16x # 0)
<=> 11/16x = x(5+16x)/(16x)
<=>11 = x(5+16x)
<=> 11 = 16x^2 + 5x
<=> x =-1 or x =11/16
17^2 - 12^2
289 - 144 = 145
Sqrt of 145 = 12.04159…
The answer rounded to two decimal places is 12.04 cm.